# Difference between revisions of "2015 AMC 10B Problems/Problem 22"

## Problem

In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? $[asy] pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("A",A,N); label("B",B,E); label("C",C,SE); label("D",D,SW); label("E",E1,W); label("F",F,NW); label("G",G,NE); label("H",H,E); label("I",I,S); label("J",J,W); [/asy]$ $\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$

## Solution

Triangle $AFG$ is isosceles, so $AG=AF=1$. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG$. Therefore, $JH=AF=1$.

Since $\triangle AJH \sim \triangle AFG$, $\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1$. From this, we get $FG=\frac{\sqrt{5} -1}{2}$.

Since $\triangle DIJ \cong \triangle AFG$, $DJ=DI=AF=1$. Since $\triangle AFG \sim ADC$, $\frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}$. Thus, $CD = (\sqrt{5} - 1) + (\frac{\sqrt{5} - 1}{2})^2 = \frac{\sqrt{5} +1}{2}$

Therefore, $FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$